Sherman-Woodbery Formula for the Inverse Matrix

Problem 250

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies
\[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\]
Define the matrix
\[A=I+\mathbf{u}\mathbf{v}^{\trans}.\]

Prove that $A$ is invertible and the inverse matrix is given by the formula
\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans},\]
where
\[a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}.\]
This formula is called the Sherman-Woodberry formula.

Proof.

Let us put
\[B=I-a\mathbf{u}\mathbf{v}^{\trans},\]
the matrix given by the Sherman-Woodberry formula.
We compute $AB$ and $BA$ and show that they are equal to the identity matrix $I$.

Let us first compute the matrix product $AB$. We have
\begin{align*}
AB&=(I+\mathbf{u}\mathbf{v}^{\trans})(I-a\mathbf{u}\mathbf{v}^{\trans})\\
&=I-a\mathbf{u}\mathbf{v}^{\trans}+\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}\\
&=I+(1-a)\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans} \tag{*}
\end{align*}
By using the defining formula for $a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}$, we have
\[a(1+\mathbf{v}^{\trans}\mathbf{u})=1,\]
and thus
\[1-a=a\mathbf{v}^{\trans}\mathbf{u}. \tag{**}\]

Note that the third term in (*) $-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}$ contains $\mathbf{v}^{\trans}\mathbf{u}$ in the middle, and $\mathbf{v}^{\trans}\mathbf{u}$ is just a number. Thus we can factor out this number and get
\begin{align*}
-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}&=-a\mathbf{u}(\mathbf{v}^{\trans}\mathbf{u})\mathbf{v}^{\trans}\\
&=-a(\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans} \tag{***}
\end{align*}
Inserting (**) and (***) into (*), it follows that we have
\begin{align*}
AB&=I+(a\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans}-a(\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans}\\
&=I.
\end{align*}
Thus we have proved $AB=I$.

Now we compute $BA$. We have
\begin{align*}
BA&=(I-a\mathbf{u}\mathbf{v}^{\trans})(I+\mathbf{u}\mathbf{v}^{\trans})\\
&=I+\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}\\
&=I+(1-a)\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}
\end{align*}
and this is exactly the expression (*), hence $BA=AB=I$.

Therefore, we conclude that the matrix $A$ is invertible and the inverse matrix is $B$. Hence
\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans}\]
and we have proved the Sherman-Woodberry formula.

Comment.

The invertible matrix theorem says that once we have $AB=I$, then we have automatically $BA=I$ and the inverse matrix of $A$ is $B$, that is, $A^{-1}=B$.
So in the above proof, after proving $AB=I$, you may conclude that $A$ is invertible and $A^{-1}=B$.

Related Question.

Problem.
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula:
\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]

Construction of a Symmetric Matrix whose Inverse Matrix is Itself
Let $\mathbf{v}$ be a nonzero vector in $\R^n$.
Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$.
Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by
\[A=I-a\mathbf{v}\mathbf{v}^{\trans},\]
where […]

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